For \(S\subset \mathbb {H}\) , let
\(\mu (S) = \lim _{\epsilon \rightarrow 0} \inf _{C_\epsilon (S)} \sum _{U\in C_\epsilon (S)} \mathcal {L}(\Pi (U))\) 
where \(C_\epsilon (S)\)  is the set of covers of \(S\)  by sets of diameter at most \(\epsilon \) . This is a Borel measure on \(\mathbb {H}\)  by [1]}, and the restriction of \(\mu \)  to any intrinsic Lipschitz graph is the pullback of \(\mathcal {L}|_{V_0}\) , i.e., \(\mu (S)=\mathcal {L}(\Pi (S))\)  for any intrinsic Lipschitz graph \(\Gamma \)  and any Borel set \(S\subset \Gamma \) . Consequently, \(\mu \)  satisfies (REF ).
By the area formula, [2]}, if \(\phi \)  is a \(\lambda \) –intrinsic Lipschitz function and \(S \subset \Gamma _\phi \)  is Borel, then
\(\mathcal {H}^3(S) \approx \int _{\Pi (S)} \sqrt{1+(\nabla _\phi \phi (v))^2} \,\mathrm {d}v = \int _{S} \sqrt{1+(\nabla _\phi \phi (v))^2} \,\mathrm {d}v \approx _\lambda \mu (S).\) 
Since, by [3]}, \(\mathcal {H}^{3}|_{\Gamma _\phi }\)  is an Ahlfors 3–regular measure this implies that \(\mu |_{\Gamma _\phi }\)  is also Ahlfors 3–regular with constants only depending on \(\lambda \) .
Finally, we check that \(\mu \)  is left-invariant. It suffices to show that \(\mathcal {L}(\Pi (g U)) = \mathcal {L}(\Pi (U))\)  for any \(g\in \mathbb {H}\)  and any Borel set \(U\subset \mathbb {H}\) . First, for any \(g, h\in \mathbb {H}\) ,
\(\Pi (gh)=gh Y^{-y(g)-y(h)}=g\left(h \cdot Y^{-y(h)}\right) Y^{-y(g)} = \Pi (g\Pi (h)).\) 
Let \(\beta _g\colon V_0\rightarrow V_0\) , \(\beta _g(v) = \Pi (g v)\) , so that \(\Pi (gU) = \beta _g(\Pi (U))\) . Let \(g=(x,y,z)\in \mathbb {H}\)  and \(v=(x^{\prime },0,z^{\prime })\in V_0\) . Then
\(\beta _g(v) = \Pi (g v) = g v Y^{-y} = \left(x + x^{\prime }, 0, z + z^{\prime } - y x^{\prime }-\frac{1}{2}xy\right).\) 
That is, \(\beta _g\)  is an affine transformation of \(V_0\)  with determinant 1. Thus \(\mathcal {L}(\Pi (gU))= \mathcal {L}(\beta _g(\Pi (U)))=\mathcal {L}(\Pi (U))\) . By (REF ), \(\mu \)  is a left-invariant measure.
